Autonomous ODE $\dot{x}=f(x)$: $\lim_{t\rightarrow\infty}x(t)=x^*\Rightarrow f(x^*)=0$ 
Let $x : [0,\infty) \to \mathbb{R}^d$ be a solution for the autonomous ODE $$\dot{x} = f(x)$$ where $f : \mathbb{R}^d \to \mathbb{R}^d$ is a Lipschitz continuous vector field. We know that $$\lim\limits_{t\to\infty}x(t)=x^*$$ where $x^*\in\mathbb{R}^d$. Show that $f(x^*)=0$.

I played around with just inserting the limit into $f$ as an argument and pulling it out (since $f$ is continuous) and I could also imagine where one would use the fact that $f$ is Lipschitz continuous. However, I have no idea where the fact that the ODE is autonomous and $x$ defined on $[0,\infty)$ is important. Hence I'm struggling with this proof.
 A: Each component $x_k$ of $x$ is a real-valued function on $[0, \infty)$, so that we can apply the mean-value theorem: For $n \in \Bbb N$
$$
x_k(n+1)-x_k(n) = \dot{x}_k(t_n) = f_k(x(t_n))
$$
for some $t_n \in (n, n+1)$. For $n \to \infty$ the left-hand side has the limit zero. On the right-hand side
$$
 t_n \to \infty \implies x(t_n) \to  x^* \implies f_k(x(t_n)) \to f_k(x^*)
$$ 
since $f$ is continuous. 
It follows that $f_k(x^*)=0$ for each component of $f$, i.e. $f(x^*) = 0$.
Remark: The Lipschitz-continuity of $f$ guarantees the existence of a solution on $[0, \infty)$, but is not needed otherwise in the above proof.
A: The statement is simply untrue for the nonautomous case. Take for example 
$$f(x,t)=-x+e^{-t}$$
The solution tends towards zero but $f(0,t)\neq0$.
A: Assume that $f(x^\ast)=V\neq 0$ and $f$ is $C$-Lipschitz
For $n$ s.t. $\frac{C}{n}< \frac{|V|}{2}$, there is $N$ s.t. $t\geq
N$ implies $|x(t)-x^\ast|< \frac{1}{n}$
Then $|f(x(t))-f(x^\ast)|<\frac{C}{n}<\frac{|V|}{2}$.
Hence $ f(x(t))=a(t)V +W(t),\ \frac{1}{2}<a(t)  < \frac{3}{2},\
 V\perp W(t)$.
Hence $$ V\cdot \{ x(t)-x(N) \}= V\cdot \int^t_N\ f(x(t)) dt \geq
\frac{1}{2}|V|^2 (t-N) $$
Hence $x(t)$ does not converge.
